dxdt+2x=0
with the initial condition x(0) = 1
and plot the solutions.
dxdt2+2dxdt+2x = e-t
Theory
Solving first order differential equation
An example of using ODEINT is with the following differential equation with parameter the initial condition x(0) =5 and the following differential equation.
dxdt+2x=0
The Python code first imports the needed Numpy, Scipy, and Matplotlib packages. The model, initial conditions, and time points are defined as inputs to ODEINT to numerically calculate x
Programe
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# function that returns dx/dt
def model(x,t):
k = 2
dxdt = -k * x
return dxdt
# initial condition
x0 = 5
# time points
t = np.linspace(0,20)
# solve ODE
x = odeint(model,x0,t)
# plot results
plt.plot(t,x,label="K=2")
plt.title("FIrst order Differential equation response")
plt.xlabel('time')
plt.ylabel('x')
plt.legend()
plt.show()
Output
An optional fourth input is args that allows additional information to be passed into the model function. The args input is a tuple sequence of values. The argument k is now an input to the model function by including an additional argument.
Programe
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# function that returns dx/dt
def model(x,t,k):
dxdt = -k * x
return dxdt
# initial condition
x0 = 5
# time points
t = np.linspace(0,20)
# solve ODE
k=.1
x1 = odeint(model,x0,t,args=(k,))
k=.5
x2 = odeint(model,x0,t,args=(k,))
k=1
x3 = odeint(model,x0,t,args=(k,))
# plot results
plt.plot(t,x1,label="K=.1")
plt.plot(t,x2,label="K=.5")
plt.plot(t,x3,label="K=1")
plt.title("First order Differential equation response with diff K values")
plt.xlabel('time -->')
plt.ylabel('x-->')
plt.legend()
plt.show()
Output
Current transient through an RC network
An RC circuit is a circuit with both a resistor (R) and a capacitor (C). RC circuits are a frequent element in electronic devices. They also play an important role in the transmission of electrical signals in nerve cells.
A capacitor can store energy and a resistor placed in series with it will control the rate at which it charges or discharges. This produces a characteristic time dependence that turns out to be exponential. The crucial parameter that describes the time dependence is the "time constant" R C
Aside from the voltage source, the RC circuit is composed of a capacitor and a resistor, we therefore assume that self-inductance is negligible. When we close the circuit, there is no inductance here so the current will just jump to the V/R value and since the capacitor is charging up and building a voltage, we can expect the current at the resistor to drop as times goes by. The differential equation we have now is the following:
By solving it, one finds:
Therefore
The current through the resistor drops exponentially, depending on a time constant tau = 1/RC. This behaviour is indeed what we find by running the simulation in Python
Program
import numpy as np
import matplotlib.pyplot as plt
# plt.style.use('ggplot')
t = np.linspace(0,1,1000)
v= 5
r = 3000 #R Value
c = 100 * 10 ** (-6) #C value
q = c*v*(1-np.exp((-1/(r*c))*t))
i = (v/r)*np.exp((-1/(r*c))*t)
plt.plot([0,t[-1]],[c*v,c*v],label='Charge peak')
plt.plot(t,q,label='Charge of the capacitor (C)')
plt.plot(t,i,label='Current (A)')
print('Tau',1/(r*c))
print('Peak current (A)',v/r)
plt.xlabel('Time (s)')
plt.title('RC circuit')
plt.legend()
plt.show()
Output
Tau 3.3333333333333335
Peak current (A) 0.0016666666666666668
With the signal 5e^tU(t)
Program
import numpy as np
import matplotlib.pyplot as plt
t = np.linspace(0,1,1000)
v = 5 * np.exp(-1*t) #DC Voltage
r = 3000 #R Value
c = 100 * 10 ** (-6) #C value
q = c*v*(1-np.exp((-1/(r*c))*t))
i = (v/r)*np.exp((-1/(r*c))*t)
plt.plot([0,t[-1]],[c*np.average(v),c*np.average(v)],label='Charge peak')
plt.plot(t,q,label='Charge of the capacitor (C)')
plt.plot(t,i,label='Current (A)')
print('Tau',1/(r*c))
print('Peak current (A)',np.average(v)/r)
plt.xlabel('Time (s)')
plt.title('RC circuit')
plt.legend()
plt.show()
Output
Tau 3.3333333333333335
Peak current (A) 0.0010536207178662611
Solution of the second order differential equation
dxdt2+2dxdt+2x = e-t
Which give
Y’’ + 2y’ + 2y = e-t
We can turn this into two first-order equations by defining a new dependent variable.
Z == y’’ → z’ + 2z + y = e-t
With initial conditions z(0) = y(0) = 0
Program
# Import the required modules
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
def dx_dt(X, t):
# Here X is a vector such that y=X[0] and z=X[1]. This function should return [y', z']
return [X[1], -2*X[1] - 1*X[0] + np.exp(-t)]
X0 = [0, 0]
t = np.linspace(0, 10, 200)
Xs = odeint(dU_dx, X0, t)
ys = Xs[:,0]
plt.xlabel("t")
plt.ylabel("y")
plt.title("second order diffrential equation response")
plt.plot(t,ys);
# print(Us)
# print(ys)
Output
Transient response of RLC circuit
A series RLC circuit , the figure show the typical RLC circuit and described by the following equation
The solution of Equation is
where Ï„ is a time constant determined by
In Equation, ω is the natural resonant frequency determined by:
Figure The transient voltage uc across capacitor for a RLC circuit
Program
#importing the library
import numpy as np
import matplotlib.pyplot as plt
L=1 # value of Inductor
R=1 # value of resistor
C=.001 #value of capacitor
tow=2*(L/R) #time constant
t = arange(0, 10, 0.01)
E1=5
E2=5*exp(-1*t)
w=sqrt((1/(L*C)-((R*R)/(4*(L*L))))) # resonant frequency
u_c1 = E1*exp(-1*(t)/tow)*cos(w*t) # response with E = 5V
u_c2 = E2*exp(-1*(t)/tow)*cos(w*t) # response with E =5E^(-t)U(t)
#Plotting the values
rcParams["figure.figsize"] = (7,10) # changing the figure size
fig, ax = plt.subplots(2)
ax[0].plot(t,u_c1)
ax[0].set_title("Transient response for E =5V")
ax[0].set_xlabel("t")
ax[0].set_ylabel("u_c")
ax[1].plot(t,u_c2)
ax[1].set_title("Transient response for E =5E^(-t)U(t)")
ax[1].set_xlabel("t")
ax[1].set_ylabel("u_c")
show()
Output
inference
dxdt+2x=0
with the initial condition x(0) = 1
and plotted the solutions.
dxdt2+2dxdt+2x = e-t
